How do I extract the matching part of a regex on Solaris

Question

GNU's grep has the option --only-matching, which prints just the matching region of a regular expression. I'm on a Solaris 5.10 box without any GNU tools installed, and I'm trying to achieve the same thing. Example:

grep -Eo "[0-9]+ ms" *.log

Is there a sed or awk expression that can do the same thing?

Answer 1

sed -n 's/.*\([0-9]\+ ms\).*/\1/p' *.log

Answer 2

Try using /usr/sfw/bin/ggrep

/usr/sfw/bin/ggrep -V grep (GNU grep) 2.5

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Answer 3

Solaris 5.10 includes bash 3.00, which has a modern regex engine. The follow command works, though it's quite a bit more verbose than a grep or sed solution.

cat *.log | while read line; 
do 
  if [[ $line =~ "([0-9]+) ms" ]]; then 
    echo "${BASH_REMATCH[1]}"; 
  fi;
done 

(Formatted on multiple lines for readability)